Question

The null and alternative hypotheses are:

H0:μd≤0H0:μd≤0

H1:μd>0H1:μd>0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

	Day
	1	2	3	4
Day shift	13	12	13	17
Afternoon shift	11	9	12	15

At the 0.10 significance level, can we conclude there are more defects produced on the Afternoon shift?

a. State the decision rule. (Round the final answer to 3 decimal places.)

H₀ is ___ if t > ___ .

b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)

Test statistic ___

c. What is your decision regarding the null hypothesis?

H₀ should ___ .

d. Determine the p-value. (Round the final answer to 4 decimal places.)

The p-value is ___.

Answer 1

Given:

The null and alternative hypotheses are:

H0 : μd ≤ 0

H1: μd > 0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

	Day
	1	2	3	4
Day shift	13	12	13	17
Afternoon shift	11	9	12	15

Significance level, = 0.10

a) Decision rule :

Reject Ho if observed test statistics > tcritical otherwise fail to reject Ho.

b) Table for calculating mean and standard deviation:

Day	X	Y	d = X-Y	d2
1	13	11	2	4
2	12	9	3	9
3	13	12	1	1
4	17	15	2	4
Total			d = 8	d2 = 18

Since t >1.638, we reject null hypothesis.

Decision: Reject null hypothesis, Ho.

Conclusion: There is sufficient evidence to conclude that there are more defects produced on the Afternoon shift.

P-value:

P-value corresponding to test statistic, t = 4.899 and degree of freedom, df = 3 is 0.0081

P-value = 0.0081